Assertion : A solid sphere and a ring of same mass and radius are released simultaneously from the top of an inclined surface. The two objects roll down the plane without slipping. They reach the bottom of the incline with equal linear speeds
Reason : Decrease in potential enery for both is the same.

Options

A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion
B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion
C. If Assertion is true but Reason is fasle
D. If Assertion is false but Reason is true

Explanation

To solve the problem, we need to analyze the assertion and reason provided regarding the motion of a solid sphere and a ring rolling down an incline.

### Step-by-Step Solution:

1. **Understanding the Scenario**:
- We have a solid sphere and a ring, both with the same mass (m) and radius (r), released from the top of an inclined plane of height (h). Both objects roll down the incline without slipping.

2. **Potential Energy at the Top**:
- The potential energy (PE) at the top for both objects is given by:
\[ PE = mgh \]
- This potential energy will convert into kinetic energy as they roll down the incline.

3. **Kinetic Energy at the Bottom**:
- The total kinetic energy (KE) for a rolling object is the sum of translational and rotational kinetic energy:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
- Here, \(I\) is the moment of inertia and \(\omega\) is the angular velocity. For rolling without slipping, we have the relation \(v = r\omega\).

4. **Moment of Inertia**:
- For the solid sphere, the moment of inertia \(I\) is:
\[ I_{\text{sphere}} = \frac{2}{5} m r^2 \]
- For the ring, the moment of inertia \(I\) is:
\[ I_{\text{ring}} = m r^2 \]

5. **Applying Energy Conservation**:
- For the solid sphere:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v}{r}\right)^2 \]
Simplifying, we find:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \frac{7}{10} mv^2 \]
Thus,
\[ v^2 = \frac{10gh}{7} \]

- For the ring:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} (m r^2) \left(\frac{v}{r}\right)^2 \]
Simplifying, we find:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} mv^2 = mv^2 \]
Thus,
\[ v^2 = \frac{gh}{1} \]

6. **Comparing Linear Speeds**:
- From the equations derived, we see that the linear speeds at the bottom of the incline are different:
- For the sphere: \(v_{\text{sphere}} = \sqrt{\frac{10gh}{7}}\)
- For the ring: \(v_{\text{ring}} = \sqrt{gh}\)

7. **Conclusion**:
- The assertion that both objects reach the bottom of the incline with equal linear speeds is **false**.
- The reason that the decrease in potential energy for both is the same is **true** because both start with the same height \(h\).

### Final Answer:
- Assertion: False
- Reason: True